2022 A/L Physics Past Paper · Question 43 (English Medium)

Answer: (2) $\dfrac{AT}{rg}$

Given information

What we know:

A cylindrical container has a small round hole of radius $r$ in its flat bottom. Its cross-sectional area is $A$.

A liquid of surface tension $T$ is poured in. When the liquid reaches a certain height $h$ above the hole, it starts to seep through.

Then the container is filled to height $\dfrac{h}{2}$. An object floats on the surface. What smallest mass $m$ makes the liquid start to seep?

We keep the letters, and later we check with sample values $A = 0.02\ \text{m}^2$, $T = 0.07\ \text{N m}^{-1}$, $r = 1 \times 10^{-4}\ \text{m}$, $\rho = 1000\ \text{kg m}^{-3}$ and $g = 10\ \text{m s}^{-2}$.

What the question asks

We want: the smallest mass $m$ of the floating object.

What each letter means:

$A$ ($\text{m}^2$): the cross-sectional area of the container.

$r$ ($\text{m}$): the radius of the hole.

$T$ ($\text{N m}^{-1}$): the surface tension of the liquid.

$h$ ($\text{m}$): the height of the liquid when it starts to seep.

$\rho$ ($\text{kg m}^{-3}$): the density of the liquid.

$g$ ($\text{m s}^{-2}$): the acceleration due to gravity.

$p$ ($\text{Pa}$): a pressure.

$m$ ($\text{kg}$): the mass of the object.

Relevant theory

What is happening:

The liquid presses down on the hole. The deeper the liquid, the bigger the pressure at the hole.

In a small hole, the liquid forms a curved skin. The surface tension of the skin can hold back a pressure up to $\dfrac{2T}{r}$.

When the pressure of the liquid at the hole gets bigger than that, the liquid seeps through.

At half the height, the pressure is half as big, which the skin can hold.

The floating object pushes down on the liquid with its weight. A floating object displaces liquid of the same weight. The container is a cylinder, so the weight $mg$ spreads over the area $A$ of the bottom. The pressure at the bottom grows by $\dfrac{mg}{A}$.

We assume the liquid does not spill and the object does not touch the sides.

The surface tension of a liquid is defined as the force acting along the surface of the liquid

on a unit length of an imaginary line drawn on the surface of the liquid, normal to and

towards one side of the line.

$p_1 - p_2 = \dfrac{2T}{r}$

Maximum excess pressure the drop can hold = $\dfrac{2T}{r_0}$ when $r_0$ is the radius of the hole.

This is the expression for the hydrostatic pressure at O.

The pressure of a liquid at a depth $h$ is $p = h\rho g$.

The book finds, from the balance of the forces on half a bubble, that a curved liquid surface of radius $r$ takes up a difference of pressure of $\dfrac{2T}{r}$ across a single surface. The liquid in the hole is held by one surface of this kind.

For example, a tea strainer. Water sits in its small holes and does not run through. It runs through when there is enough water on top, or when you press down on the water with a spoon.

The strainer holds back the water in the same way: the thin skin in each hole holds the pressure back until it is too big.

Required equations

$h\rho g = \dfrac{2T}{r}$

$\dfrac{h}{2}\rho g = \dfrac{T}{r}$

$\dfrac{T}{r} + \dfrac{mg}{A} = \dfrac{2T}{r}$

$m = \dfrac{AT}{rg}$

Applying it to this question

$\dfrac{AT}{rg}$ has the unit $\dfrac{\text{m}^2 \times \text{N m}^{-1}}{\text{m} \times \text{m s}^{-2}} = \dfrac{\text{N s}^2}{\text{m}} = \text{kg}$, so the answer is a mass.

$1400$

$\dfrac{2T}{r} = \dfrac{2 \times 0.07}{1 \times 10^{-4}}$

$0.14$

$h = \dfrac{1400}{\rho g} = \dfrac{1400}{1000 \times 10}$

$700$

$\dfrac{h}{2}\rho g = 0.07 \times 1000 \times 10$

$1400 - 700$

$1.4$

$m = \dfrac{700 \times A}{g} = \dfrac{700 \times 0.02}{10}$

Reasoning and calculation

The extra pressure is $\dfrac{mg}{A} = \dfrac{T}{r}$. So $m = \dfrac{AT}{rg}$.

That is option (2).

Checks

$a * t / (r * g)$

Does it make sense?

The mass is bigger for a wider container, because the weight is spread over a bigger area. It is bigger for a bigger surface tension, because the skin holds more. It is smaller for a bigger hole, because the skin in a wider hole holds back less.

If the container were filled to the full height $h$ there would be no need for an object: $m = 0$. At half the height, the object has to make up half of the pressure, which is $\dfrac{T}{r}$.

Conclusion

$m = \dfrac{AT}{rg}$

The smallest mass of the object is $m = \dfrac{AT}{rg}$.

The answer is (2).

Write the pressure that makes the liquid seep, $\dfrac{2T}{r}$. Find how much the liquid gives at the new height, $h\rho g$. The object has to add the difference, and its weight spread over the area $A$ gives $\dfrac{mg}{A}$.

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